Question

A 2.54 \times 10^{-2}-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 115 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 9.42 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

Name: A 2.54 ×10^-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 115 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 9.42 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?
Uploaded: 2022-01-30T19:22:29-08:00
Duration: 2 min 24 s
Channel: Rashmi Prakash
Description: A 2.54 ×10^-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 115 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 9.42 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

          A 2.54 \times 10^{-2}-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 115 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 9.42 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

A 2.54 ×10^-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 115 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 9.42 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

Added by Derrick M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Rashmi Prakash

01/30/2022

Step 1

The potential energy stored in a spring is given by the formula: Potential energy = (1/2) * k * x^2 where k is the spring constant and x is the displacement from the equilibrium position. In this case, the spring is initially unstrained, so the displacement is Show more…

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Texts: A 2.54x10^-2 kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 115 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 9.42 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion? Number Units